Chapter 22, Problem 16P

(a)

To determine

The frequency of the proton.

Answer to Problem 16P

The frequency of the proton is. $7.66\times {10}^7{\text{s}}^{-1}$.

Explanation of Solution

Given Info: The accelerating voltage is $600\text{V}$, the outer radius is $0.350\text{m}$ and the magnitude of magnetic field is $0.800\text{T}$.

The formula for the frequency is,

$f=\frac{qB}{2\pi m}$

Here,

$B$ is the magnitude of magnetic field.

$m$ is the mass of proton.

$q$ is the charge of proton.

$V$ is the accelerating voltage.

Substitute $0.800\text{T}$ for $B$, $1.6\times {10}^{-27}\text{kg}$ for $m$, $1.6\times {10}^{-19}\text{C}$ for $q$ and $600\text{V}$ for $V$ in above equation to find $f$.

$\begin{array}{c}f=\frac{\left(1.6\times {10}^{-19}\text{C}\right)\left(0.800\text{T}\right)}{2\pi \left(1.6\times {10}^{-27}\text{kg}\right)}\\ =7.66\times {10}^7{\text{s}}^{-1}\end{array}$

Thus, the frequency of the proton is. $7.66\times {10}^7{\text{s}}^{-1}$.

Conclusion:

Therefore, the frequency of the proton is. $7.66\times {10}^7{\text{s}}^{-1}$.

(b)

To determine

The exit speed of the proton.

Answer to Problem 16P

The exit speed of the proton is. $2.68\times {10}^7\text{m}/\text{s}$.

Explanation of Solution

Given Info: The accelerating voltage is $600\text{V}$, the outer radius is $0.350\text{m}$ and the magnitude of magnetic field is $0.800\text{T}$.

The formula for the speed is,

$v=\frac{qBr}{m}$

Here,

$r$ is the radius of the outermost orbit.

Substitute $0.800\text{T}$ for $B$, $1.6\times {10}^{-27}\text{kg}$ for $m$, $1.6\times {10}^{-19}\text{C}$ for $q$ and $0.350\text{m}$ for $r$ in above equation to find $v$ .

$\begin{array}{c}v=\frac{\left(1.6\times {10}^{-19}\text{C}\right)\left(0.800\text{T}\right)\left(0.350\text{m}\right)}{\left(1.6\times {10}^{-27}\text{kg}\right)}\\ =2.68\times {10}^7\text{m}/\text{s}\end{array}$

Thus, the exit speed of the proton is. $2.68\times {10}^7\text{m}/\text{s}$.

Conclusion:

Therefore, the exit speed of the proton is. $2.68\times {10}^7\text{m}/\text{s}$.

(c)

To determine

The maximum kinetic energy.

Answer to Problem 16P

The maximum kinetic energy is $3.75\text{MeV}$.

Explanation of Solution

Given Info: The accelerating voltage is $600\text{V}$, the outer radius is $0.350\text{m}$ and the magnitude of magnetic field is $0.800\text{T}$.

The formula for the kinetic energy is,

$K{E}_{\max }=\frac{1}{2}m{v}^2$

Substitute $1.67\times {10}^{-27}\text{kg}$ for $m$ and $2.68\times {10}^7\text{m}/\text{s}$ for $v$ in above equation to find $K{E}_{\max }$.

$\begin{array}{c}K{E}_{\max }=\frac{1}{2}\left(1.67\times {10}^{-27}\text{kg}\right){\left(2.68\times {10}^7\text{m}/\text{s}\right)}^2\\ =5.99\times {10}^{-13}\text{J}\times \frac{\left(6.2\times {10}^{12}\text{MeV}\right)}{1\text{J}}\\ =3.75\text{MeV}\end{array}$

Thus, the maximum kinetic energy is $3.75\text{MeV}$.

Conclusion:

Therefore, the maximum kinetic energy is $3.75\text{MeV}$.

(d)

To determine

The number of revolutions.

Answer to Problem 16P

The number of revolutions are $3.13\times {10}^3\text{revolutions}$.

Explanation of Solution

Given Info: The accelerating voltage is $600\text{V}$, the outer radius is $0.350\text{m}$ and the magnitude of magnetic field is $0.800\text{T}$.

The formula for the number of revolutions is,

$N=\frac{E_{\text{max}}}{2qV}$

Substitute $600\text{V}$ for $V$, $1.6\times {10}^{-19}\text{C}$ for $q$ and $4.4\times {10}^{-13}\text{J}$ for $K{E}_{\max }$ in above equation to find $N$ .

$\begin{array}{c}N=\frac{\left(4.4\times {10}^{-13}\text{J}\right)}{2\left(1.6\times {10}^{-19}\text{C}\right)\left(600\text{V}\right)}\\ =3.13\times {10}^3\text{revolutions}\end{array}$

Thus, the number of revolutions are $3.13\times {10}^3\text{revolutions}$.

Conclusion:

Therefore, the number of revolutions are $3.13\times {10}^3\text{revolutions}$.

(e)

To determine

The time of acceleration.

Answer to Problem 16P

The time of acceleration is $2.57\times {10}^{-4}\text{s}$.

Explanation of Solution

Given Info: The accelerating voltage is $600\text{V}$, the outer radius is $0.350\text{m}$ and the magnitude of magnetic field is $0.800\text{T}$.

The formula for the number of revolutions is,

$T=\frac{2\pi m}{qB}$

Substitute $0.800\text{T}$ for $B$, $1.6\times {10}^{-27}\text{kg}$ for $m$ and , $1.6\times {10}^{-19}\text{C}$ for $q$ in above equation to find $T$ .

$\begin{array}{c}T=\frac{2\pi \left(1.6\times {10}^{-27}\text{kg}\right)}{\left(1.6\times {10}^{-27}\text{kg}\right)\left(0.800\text{T}\right)}\\ =2.57\times {10}^{-4}\text{s}\end{array}$

Thus, the time of acceleration is $2.57\times {10}^{-4}\text{s}$.

Conclusion:

Therefore, the time of acceleration is $2.57\times {10}^{-4}\text{s}$.